Answer to Question #147279 in Mechanical Engineering for Sebastian

Question #147279
A car traveling on a rough and bumpy road, which is modeled with the basic displacement y (t) = (0.01) sin (5.818t) m. The car suspension system, which is modeled with 1-degree of freedom, has a rigidity of k = 4 x 10^5 N/m, and a damping coefficient of c = 40 x 10^3 kg/s, and the mass of the car m = 1007 kg. Calculate the absolute displacement amplitude of the mass of the car.
Expert's answer

Here the displacement equation is given as

y(t)=0.01 sin(5.818 t)

here degree of freedom of car is 1, and stiffness is "4\\times 10^5 N\/m"

and damping coefficient is , c= "4\\times 10^3 kg\/s" and mass of car is 1007 kg

"\\omega _n= \\sqrt \\frac{k}{m}=\\sqrt{\\frac{4\\times 10^5}{1007}}=19.93"

and damping coefficient is

"\\zeta=\\frac{c}{2\\sqrt{mk}}=\\frac{4\\times 10^3}{2\\sqrt{1007\\times 4\\times 10^5}}=0.9965"

Now the frequncy ratio we are getting value of frequency from equation as , "\\omega=5.818"

"r=\\frac{ \\omega}{\\omega_n}=\\frac{5.818}{19.93}=0.292"

And in case of single degree of freedom for damped motion we have amplitude formula as

"X=Y\\sqrt{ \\frac{ 1+(2\\zeta r)^2}{(1+r^2)^2 +(2 \\zeta r)^2}}"

"X=0.01\\sqrt{ \\frac{ 1+(2\\times0.997\\times 0.292)^2}{(1+0.292^2)^2 +(2 \\times0.997\\times 0.292)^2}}=0.011 m"

So, magnitude of absolute displacement is 0.011 m

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